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To solve this problem, we need to find the shortest distance between the given pair of lines:

\[
\frac{x-1}{2} = \frac{y+1}{3} = z
\]
and
\[
\frac{x+1}{5} = \frac{y-2}{1}; \quad z = 2
\]

### Step-by-Step Approach:

1. **Equation of the First Line:**
   \[
   \frac{x-1}{2} = \frac{y+1}{3} = z
   \]
   This can be rewritten in parametric form by setting \( z = t \) (a parameter):
   \[
   x = 2t + 1, \quad y = 3t - 1, \quad z = t
   \]
   So, the parametric equation of the first line is:
   \[
   (x, y, z) = (2t+1, 3t-1, t)
   \]

2. **Equation of the Second Line:**
   \[
   \frac{x+1}{5} = \frac{y-2}{1}; \quad z = 2
   \]
   From the first part, we get:
   \[
   x = 5s - 1, \quad y = s + 2
   \]
   and since \( z = 2 \) is constant for this line, the parametric form is:
   \[
   (x, y, z) = (5s - 1, s + 2, 2)
   \]

3. **Direction Vectors:**
   - For the first line: The direction vector is obtained from the parametric form:
     \[
     \vec{d_1} = \langle 2, 3, 1 \rangle
     \]
   - For the second line: The direction vector is:
     \[
     \vec{d_2} = \langle 5, 1, 0 \rangle
     \]

4. **Shortest Distance Formula:**
   The shortest distance \( d \) between two skew lines can be found using the formula:
   \[
   d = \frac{|(\vec{r_2} - \vec{r_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}
   \]
   where \( \vec{r_1} \) and \( \vec{r_2} \) are points on the two lines, and \( \vec{d_1} \) and \( \vec{d_2} \) are their direction vectors.

5. **Finding \( \vec{r_1} \) and \( \vec{r_2} \):**
   - A point on the first line (when \( t = 0 \)) is \( \vec{r_1} = (1, -1, 0) \).
   - A point on the second line (when \( s = 0 \)) is \( \vec{r_2} = (-1, 2, 2) \).

6. **Calculating the Cross Product \( \vec{d_1} \times \vec{d_2} \):**
   \[
   \vec{d_1} = \langle 2, 3, 1 \rangle, \quad \vec{d_2} = \langle 5, 1, 0 \rangle
   \]
   The cross product is:
   \[
   \vec{d_1} \times \vec{d_2} = \langle 3 \times 0 - 1 \times 1, 1 \times 5 - 2 \times 0, 2 \times 1 - 3 \times 5 \rangle = \langle -1, 5, -13 \rangle
   \]

7. **Finding \( \vec{r_2} - \vec{r_1} \):**
   \[
   \vec{r_2} - \vec{r_1} = \langle -1 - 1, 2 - (-1), 2 - 0 \rangle = \langle -2, 3, 2 \rangle
   \]

8. **Dot Product:**
   Now, calculate the dot product:
   \[
   (\vec{r_2} - \vec{r_1}) \cdot (\vec{d_1} \times \vec{d_2}) = \langle -2, 3, 2 \rangle \cdot \langle -1, 5, -13 \rangle
   \]
   \[
   = (-2)(-1) + (3)(5) + (2)(-13) = 2 + 15 - 26 = -9
   \]

9. **Magnitude of the Cross Product:**
   \[
   |\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + 5^2 + (-13)^2} = \sqrt{1 + 25 + 169} = \sqrt{195}
   \]

10. **Shortest Distance:**
    Finally, the shortest distance is:
    \[
    d = \frac{| -9 |}{\sqrt{195}} = \frac{9}{\sqrt{195}} = \frac{9}{\sqrt{195}} \times \frac{\sqrt{195}}{\sqrt{195}} = \frac{9\sqrt{195}}{195} = \frac{3\sqrt{195}}{65}
    \]

Thus, the shortest distance between the two lines is \( \frac{3\sqrt{195}}{65} \).

---

Would you like further details on any of the steps, or do you have any questions?

Here are five related questions for deeper understanding:
1. How is the cross product of two vectors calculated geometrically?
2. What is the geometric interpretation of the shortest distance between skew lines?
3. How do we parametrize the equation of a line in 3D space?
4. What are some practical applications of finding the distance between lines?
5. How do direction vectors relate to the orientation of lines?

**Tip:** The shortest distance between two skew lines can always be found using vector methods involving the cross product and dot product, which are foundational operations in vector calculus.

Find the shortest distance between the pair of lines (x-1)/2 = (y+1)/3 = z and (x+1)/5 = (y-2)/1; z = 2.

Learn how to calculate the shortest distance between two skew lines using parametric equations and vector methods. This example involves 3D geometry concepts such as direction vectors, cross product, and dot product, suitable for high school and undergraduate students.

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